JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the function \(f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6\) have a maxima for some value of \(x < 0\) and a minima for some value of \(x > 0\). Then, the set of all values of \(p\) is \(......\)
- A \(\left(\frac{9}{2}, \infty\right)\)
- B \(\left(0, \frac{9}{2}\right)\)
- C \(\left(-\infty, \frac{9}{2}\right)\)
- D \(\left(-\frac{9}{2}, \frac{9}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(-\infty, \frac{9}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6\) \(f^{\prime}(x)=6 x^2+2(2 p-7) x+3(2 p-9)\) \(f^{\prime}(0) < 0\) \(\therefore 3(2 p-9) < 0\) \(\quad p < \frac{9}{2}\) \(\quad p \in\left(-\infty, \frac{9}{2}\right)\)
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